Question

For a reaction \(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{P}\), when concentration of \(\mathrm{B}\) alone is doubled, \(\frac{\mathrm{t}_{1}}{2}\) does not change and when concentrations of both \(A\) and \(B\) is doubled, rate increases by a factor of 4 . The unit of rate constant is,
(A) \(\mathrm{s}^{-1}\)
(B) \(\mathrm{L} \mathrm{mol}^{-1} \mathrm{~S}^{-1}\)
(C) \(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\)
(D) \(\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}\)

Answer 1

Ans: (B)

Rate law: \(r=k[A]^{\alpha}[B]^{\beta} \quad B\) follows first order kinetics. \(\therefore \beta=1\)
\(\frac{r_{2}}{r_{1}}=\frac{[2 A]^{\alpha}[2 B]^{\beta}}{[A]^{\alpha}[B]^{\beta}}\)
or, \(4=2^{\alpha+\beta}\)
or, \(\alpha+\beta=2\)
or, \(\alpha=1\)
\(\therefore\) overall order of reaction \(=2\)
Hence unit of rate constant \(=\) L. \(\mathrm{mol}^{-1} \cdot \mathrm{S}^{-1}\)

Answer 2

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Answer 3

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